Recent content by Opalg

If n^2 + 3 = k(2n+4) then n^2 - 2kn + (3-4k) = 0. The discriminant of that quadratic must be a perfect square, say k^2 + 4k-3 = m^2. Then (k+2)^2 - 7 = m^2. The only squares that differ by 7 are 9 and 16, so k+2 = \pm4. Thus either k = 2 and n^2 - 4n - 5 = 0, giving n = -1 \text{ or }5; or k= -6...

It seems that the PF LaTeX system does not recognise the TikZ environment that is used in many MHB threads. See for example here or here.

As in a previous problem, here is an algebraic solution. I would like to see a geometric solution giving more insight into why this result holds.

Brute force solution. :censored: 🥊

Here is an outline of a solution using coordinate geometry. [scale=1.5] \coordinate [label=above right:{\textcolor{blue}O}] (O) at (0,0) ; \coordinate [label=above right:{\textcolor{blue}A}] (A) at (0,3.46) ; \coordinate [label=above right:{\textcolor{blue}B}] (B) at (1.73,0.46) ; \coordinate...

I think this result must be false. Start with the matrix $A = \begin{bmatrix}0&0&\alpha \\ \alpha&0&0 \\ 0&\alpha&0\end{bmatrix}$. Then$A^2 = \begin{bmatrix} 0&\alpha^2&0 \\ 0&0&\alpha^2 \\ \alpha^2&0&0 \end{bmatrix}$ and $A^3 = \begin{bmatrix}\alpha^3&0&0 \\ 0&\alpha^3&0 \\ 0&0&\alpha^3...

It's obvious from the graph at #1 that $S_n(x)$ increases very rapidly to $1$ as $n$ increases. So it seemed helpful to write $S_n(x)$ in the form $1$ - ?. I found that $$S_1(x) = 1 - \frac12\cos^2\left(\frac{\pi x}2\right),$$ $$S_2(x) = 1 - \frac14\cos^2\left(\frac{\pi...

Use the result $\sin x = 2\sin\frac x2\cos\frac x2$ repeatedly to see that $$\begin{aligned}\sin^2(\pi x) &= 2^2\sin^2\left(\frac{\pi x}2\right)\cos^2\left(\frac{\pi x}2\right) \\ &= 2^4\sin^2\left(\frac{\pi x}4\right)\cos^2\left(\frac{\pi x}4\right)\cos^2\left(\frac{\pi x}2\right) \\ & \vdots...

Alternatively, \sup_{x\in B_j} --> $\displaystyle \sup_{x\in B_j}$. (The symbol for sup then appears in Roman type rather than italics.)

For every $x\in B$, $f(x) \leqslant \sup_B f$ and $g(x) \leqslant \sup_B g$. Therefore $(fg)(x) = f(x)g(x) \leqslant \sup_B f \sup_B g$. Now take the sup over $B$ to get $\sup_B fg \leqslant \sup_B f \sup_B g$. A similar argument shows that $\inf_B fg \geqslant \inf_B f \inf_B g$ and so...

If the point $c$ moves so that the sum of its lengths to $a$ and $b$ is constant, then its locus will be an ellipse (the red curve in the diagram) with one focus at $a$ and the other one at $b$. The vertical line through $c$ will bisect the angle between the blue lines through $c$ at the point...

The matrix $A$ has the interesting property that its square is minus the identity matrix: $A^2 = \begin{bmatrix}-1&0\\0&-1\end{bmatrix}$. That suggests that it should correspond to the complex number $i$. In fact, $K$ consists of all matrices of the form $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$...

{"version":7,"graph":{"viewport":{"xmin":-26.663023386989984,"ymin":-12.46821832312099,"xmax":16.18142239370036,"ymax":23.98932209548482}},"randomSeed":"79bd9e65b9a1a48586c1677c8d4b0ba1","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"x^{2}\\ +\\ y^{2}\\ -\\ 12x\\...

Use the \setminus symbol: $B\setminus B'$.